Does the sequence of functions $$f_n(x) = (1+x^n)^\frac{1}{n}, \qquad x \geq 0$$ converge? If so, is the convergence uniform?
Here I figured that yes it converges pointwise to $f$ given piecewise by $f(x) = 1$ for $x \in [0,1)$ and $f(x) = x$ for $x \geq 1$. To do this my strategy was to bound above $(1+x^n)^\frac{1}{n}$ by $2^\frac{1}{n}$ for $x \in [0,1)$ and $(2x^n)^\frac{1}{n}$ for $x \geq 1$ and then taking limit as $n \to \infty$ of $2^\frac{1}{n}-1$ or $(2x^n)^\frac{1}{n}-x$ to get $0$ for both cases (with $x$ being fixed).
The part I am stuck is whether the convergence is uniform. While I noticed I can use the upper bound above for $x \in [0,1)$, I cannot for $x \geq 1$. How can I do it instead? Since I'm struggling to simplify $(1+x^n)^\frac{1}{n}$ in other ways.
Note: I cannot use calculus techniques like L'Hôpital, Taylor expansion etc., nor techniques in measure theory.
By the mean value theorem $$ \frac{1}{n}\frac{1}{{(1 + t)^{1 - 1/n} }} < (1 + t)^{1/n} - t^{1/n} < \frac{1}{n}\frac{1}{{t^{1 - 1/n} }} $$ for any $t\geq 1$. Thus with $t=x^n \geq 1$, we have $$ 0 < \frac{1}{n}\frac{1}{{(1 + x^n )^{1 - 1/n} }} < (1 + x^n )^{1/n} - x < \frac{1}{n}\frac{1}{{(x^n )^{1 - 1/n} }} \le \frac{1}{n}. $$ This shows that the convergence is uniform for $x\geq 1$.